Question: Find the inverse of the matrix, $\text B = \left[\begin{array}{rr}7 & 9 \\ 1 & 1\end{array}\right]$. Non-integers should be given either as decimals or as simplified fractions. $ B^{-1}=$
Explanation: The Strategy To find the inverse of an invertible matrix, we can use Gaussian Elimination. To do this, we do the following. First, we append the matrix $\text B$ with the identity matrix $\text I$ to get $\left[\begin{array}{c|c} \text B &\text I\end{array}\right]$. Next, we use Gaussian Elimination to reduce $\text B$ to the identity matrix, $\text I$. Performing the same operations on $\text I$ will convert it to $\text B^{-1}$, so that our new matrix becomes $\left[\begin{array}{c|c} \text I &\text B^{-1}\end{array}\right]$. Appending $\text B$ with $\text I$ $\left[\begin{array}{c|c} \text B &\text I\end{array}\right]=\left[\begin{array}{rr}7 & 9 & 1 & 0 \\ 1 & 1 & 0 & 1 \end{array}\right]$ Eliminating the leading term in the second row We want the first term of $R_2$ to equal $0$, so we subtract $\dfrac{1}{7}R_1$ from $R_2$. $\left[\begin{array}{rr}7 & 9 & 1 & 0 \\ {1} & {1} & {0} & {1} \end{array}\right]\xrightarrow{R_2-\frac{1}{7}R_1\rightarrow R_2}\left[\begin{array}{rr}7 & 9 & 1 & 0 \\ {0} & {-\dfrac{2}{7}} & {-\dfrac{1}{7}} & {1} \end{array}\right]$ Reducing the leading terms and back-solving Now, let's reduce the leading term of $R_2$ to equal $1$. $\left[\begin{array}{rr}7 & 9 & 1 & 0 \\ {0} & {-\dfrac{2}{7}} & {-\dfrac{1}{7}} & {1} \end{array}\right]\xrightarrow{-\frac{7}{2}R_2\rightarrow R_2}\left[\begin{array}{rr}7 & 9 & 1 & 0 \\ {0} & {1} & {\dfrac{1}{2}} & {-\dfrac{7}{2}} \end{array}\right]$ We are ready to back-solve to get $\left[\begin{array}{c|c} \text I &\text B^{-1}\end{array}\right]$. $\begin{aligned}\left[\begin{array}{rr}{7} & {9} & {1} & {0} \\ 0 & 1 & \dfrac{1}{2} & -\dfrac{7}{2} \end{array}\right]\xrightarrow{R_1-9R_2\rightarrow R_1} &\left[\begin{array}{rr}{7} & {0} & {-\dfrac{7}{2}} & {\dfrac{63}{2}} \\ 0 & 1 & \dfrac{1}{2} & -\dfrac{7}{2} \end{array}\right] \xrightarrow{\frac{1}{7}R_1\rightarrow R_1}\left[\begin{array}{rr}{1} & {0} & {-\dfrac{1}{2}} & {\dfrac{9}{2}} \\ 0 & 1 & \dfrac{1}{2} & -\dfrac{7}{2} \end{array}\right]\end{aligned}$ Therefore $\text B^{-1}=\left[\begin{array}{rr} -\dfrac{1}{2} & \dfrac{9}{2} \\ \dfrac{1}{2} & -\dfrac{7}{2} \end{array}\right]$. Summary $\text B^{-1}=\left[\begin{array}{rr} -\dfrac{1}{2} & \dfrac{9}{2} \\ \dfrac{1}{2} & -\dfrac{7}{2} \end{array}\right]$